If diagonals of a rhombus are 16 cm and 30 cm. What is the perimeter (in cm) of the rhombus?

The correct option is 68.

Let the rhombus be $ABCD$ where the diagonals intersect at point $O$.
Let the length of the first diagonal be $d_1=16\text{cm}$ and the length of the second diagonal be $d_2=30\text{cm}$.

We know that the diagonals of a rhombus bisect each other at right angles ($90°$).
Therefore, the lengths of the halves of the diagonals are:

$OA=\frac{d_1}{2}=\frac{16}{2}=8\text{cm}$

$OB=\frac{d_2}{2}=\frac{30}{2}=15\text{cm}$

Since the diagonals intersect at a right angle, triangle $AOB$ is a right-angled triangle with the right angle at $O$.
By applying the Pythagorean theorem to triangle $AOB$, we can find the side length $a$ (which is the hypotenuse $AB$) of the rhombus:

$A{B}^2=O{A}^2+O{B}^2$

$a^2=8^2+{15}^2$

$a^2=64+225$

$a^2=289$

$a=\sqrt{289}=17\text{cm}$

Since all four sides of a rhombus are equal in length, the perimeter is given by:

$\text{Perimeter}=4\times a$

$\text{Perimeter}=4\times 17=68\text{cm}$

Thus, the perimeter of the rhombus is 68 cm.